Conquering the Monstrous Partial Fraction: Solving Complex Equations

In summary, the problem is that the denominator factors in a complex form and I need to change them to a real form to take advantage of a known Fourier transform.
  • #1
WolfOfTheSteps
138
0

Homework Statement



I know how to use the method of partial fractions in most circumstances, but I'm working on a problem that has gotten the best of me. How do I get from the left side of the following identity to the right side?

[tex]
\frac{-2-2\omega^2}{-\omega^2+\sqrt{2}i\omega+1}
\ = \
\ 2 \ + \
\frac{-\sqrt{2}-2\sqrt{2}i}{i\omega -
\frac{-\sqrt{2}+i\sqrt{2}}{2}} \ + \
\frac{-\sqrt{2}-2\sqrt{2}i}{i\omega -
\frac{-\sqrt{2}-i\sqrt{2}}{2}}
[/tex]​

The Attempt at a Solution



I was able to factor the denominator and write the following equation:

[tex]
A\left(\frac{\sqrt{2}+i\sqrt{2}}{2} +i\omega\right) \ + \
B\left(\frac{\sqrt{2}-i\sqrt{2}}{2} +i\omega\right) \ = \
-2-2\omega^2
[/tex]​

but couldn't get much further because A and B don't have [itex]\omega^2[/itex] multiples to match up with the [itex]-2\omega^2[/itex] on the right side of the equation.

How do I handle this monster?

Note: [itex]i[/itex] is the imaginary unit.

Thanks!
 
Physics news on Phys.org
  • #2
I would like you to show how you arrived at what you give since that is not what I get.

The first thing I did was multiply both numerator and denominator by -1 to get
[tex]\frac{2+2\omega^2}{\omega^2-\sqrt{2}i\omega-1}[/tex]
Then the denominator factors as
[tex]\omega^2- \sqrt{2}i\omega- 1= (\omega-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})(\omega+\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})[/tex]
Putting those factors into the denominators and multiplying through by them gives
[tex]A(\omega+\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})+ B(\omega-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})= 2+ 2\omega^2[/tex]

Now set
[tex]\omega= \frac{\sqrt{2}}{2}+ i\frac{\sqrt{2}}{2}[/tex]
and
[tex]\omega= \frac{\sqrt{2}}{2}- i\frac{\sqrt{2}}{2}[/tex]
and it should be easy.
 
  • #3
HallsofIvy said:
I would like you to show how you arrived at what you give since that is not what I get.

Thanks for your help!

Your denominator factors are not in the form I need them to be. I didn't check it, but I assume your work is correct. The problem is I need the denominator factors to be in the [itex](a +i\omega)[/itex] (where [itex]a[/itex] is complex or real) form to take advantage of a known Fourier transform.

Here is how I arrived at my equation:

Let

[tex](A+i\omega)(B+i\omega) = 1 + \sqrt{2}i\omega-\omega^2[/tex]

Then we want A and B to satisfy the following conditions:

[tex]A+B = \sqrt{2}[/tex]
[tex]AB = 1[/tex]

Solving that system we get:

[tex]A= \frac{\sqrt{2}}{2}(1-i)[/tex]
[tex]B= \frac{\sqrt{2}}{2}(1+i)[/tex]

Hence,

[tex]1+\sqrt{2}i\omega - \omega^2 = \left(\frac{\sqrt{2}}{2}(1-i)+i\omega\right)
\left(\frac{\sqrt{2}}{2}(1+i)+i\omega\right)[/tex]

And from that I was able to get the equation I gave before:

[tex]
A\left(\frac{\sqrt{2}+i\sqrt{2}}{2} +i\omega\right) \ + \
B\left(\frac{\sqrt{2}-i\sqrt{2}}{2} +i\omega\right) \ = \
-2-2\omega^2
[/tex]

And that's where I was stuck.
 
Last edited:
  • #4
HallsofIvy,

I got it now. I found out that I have to first do long division to get the numerator to be 1 degree less than the denominator, and then use the method of partial fractions.

Thanks for your help!
 
  • #5
Dang, didn't even occur to me!
 

Related to Conquering the Monstrous Partial Fraction: Solving Complex Equations

What is "Partial Fraction From Hell"?

"Partial Fraction From Hell" is a mathematical concept that involves breaking down a complex rational function into simpler fractions. It is often considered difficult because it involves a lot of algebraic manipulation and can be time-consuming.

Why is it called "Partial Fraction From Hell"?

The term "hell" is often used metaphorically to describe something that is difficult, frustrating, or seemingly impossible. In the context of mathematics, "Partial Fraction From Hell" is used to emphasize the challenging nature of this concept.

What is the purpose of using partial fractions?

The main purpose of using partial fractions is to simplify a complex rational function into smaller, more manageable fractions. This can make solving the function easier and help in understanding the behavior of the function.

What are the key steps in solving a "Partial Fraction From Hell" problem?

The key steps in solving a "Partial Fraction From Hell" problem include identifying the type of rational function, decomposing the function into partial fractions, setting up and solving equations to determine the unknown coefficients, and finally combining the partial fractions to get the original function.

How can I improve my skills in solving "Partial Fraction From Hell" problems?

To improve your skills in solving "Partial Fraction From Hell" problems, it is important to have a strong understanding of algebraic manipulation, basic calculus, and complex numbers. Practice and patience are also key in mastering this concept.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
779
  • Calculus and Beyond Homework Help
Replies
6
Views
841
  • Calculus and Beyond Homework Help
Replies
3
Views
865
Replies
9
Views
808
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
20
Views
582
Replies
12
Views
481
  • Calculus and Beyond Homework Help
Replies
6
Views
267
  • Calculus and Beyond Homework Help
Replies
4
Views
828
  • Calculus and Beyond Homework Help
Replies
3
Views
863
Back
Top